The lifespans of porcupines in a particular zoo are normally distributed. The average porcupine lives $19$ years; the standard deviation is $1.7$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a porcupine living less than $24.1$ years.
$19$ $17.3$ $20.7$ $15.6$ $22.4$ $13.9$ $24.1$ $99.7\%$ $0.15\%$ $0.15\%$ We know the lifespans are normally distributed with an average lifespan of $19$ years. We know the standard deviation is $1.7$ years, so one standard deviation below the mean is $17.3$ years and one standard deviation above the mean is $20.7$ years. Two standard deviations below the mean is $15.6$ years and two standard deviations above the mean is $22.4$ years. Three standard deviations below the mean is $13.9$ years and three standard deviations above the mean is $24.1$ years. We are interested in the probability of a porcupine living less than $24.1$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the porcupines will have lifespans within 3 standard deviations of the average lifespan. The remaining $0.3\%$ of the porcupines will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({0.15\%})$ will live less than $13.9$ years and the other half $({0.15\%})$ will live longer than $24.1$ years. The probability of a particular porcupine living less than $24.1$ years is ${99.7\%} + {0.15\%}$, or $99.85\%$.